Name: KEY
Date
Even MORE Quadratic Applications
s(t) = –gt2 + v0t + h0
1) You and your friend decide to get celebrate the end of the first semester by going to the roof of a
twelve-story building that looks over the edge of a reflecting pool 160 feet below. You drop your math
book over the edge at the same instant that your friend throws his book straight down at 48 feet per
second. By how many seconds does his book beat yours into the water? Slower book: root 10
seconds, faster 2 seconds
The initial launch heights will be the same: you're both launching from 160 feet above ground. And the gravity
number, since you're working in feet, will be 16. Your initial velocity is zero, since you just dropped your book, but
your friends velocity is a negative 48, the negative coming from the fact that he threw his book down rather than
up. So our "height" equations are:
2
Yours: s(t) = –16t + 160
2
Friends: s(t) = –16t – 48t
+ 160
In each case, You need to find the time for the books to reach a height of zero ("zero" being "ground level"), so:
2
2
Yours: 0 = –16t + 160, t – 10 = 0, so t = ± sqrt(10)
2
2
Friends: 0 = –16t – 48t + 160, t + 3t – 10 = 0, (t + 5)(t
– 2) = 0, so t = –5 or t = 2
I will ignore the negative time values. Your friends book hits the water after two seconds, and yours hits after
sqrt(10) seconds, or after about 3.16 seconds. That is:
Your friends book hits the water about
1.16 seconds sooner than yours does.
2) The International Space Agency has finally landed a robotic explorer on an extra-solar planet. Some
probes are extended from the lander's body to conduct various tests. To demonstrate the crushing
weight of gravity on this planet, the lander's camera is aimed at a probe's ground-level ejection port,
and the port launches a baseball directly upwards at 147 feet per second (ft/s), about the top speed of a
professional pitcher. The force due to gravity on this planet is 98 ft/s2. Assuming no winds and that the
probe can scurry out of the way in time;
A) How long will it take for the ball to smack back into the surface? 3 seconds
2
Because the gravity is 98, .5(g ) = 49. The equation we will use is -49t + 147t (h0 = 0)
2
-49t + 147t = 0  -49t(t – 3). The two solutions are 0 and 3. 0 is the start time, so 3 is the answer
B) If the same experiment were conducted on Earth, how much longer would it take the ball to hit the
surface?
On Earth the force of gravity in ft/sec is 32. So (1/2)g = 16. The equation we use is
2
-16t + 147t = 0  -16t(t – 147/16). On Earth it hits the ground in 147/16 seconds (about 9.18 seconds). So
on Earth it would take about 6.18 seconds longer to hit the ground.
C) On which planet is the balls maximum height greater and by how much?
On the extra solar planet the vertex is at (1.5, 110.5)
On Earth the vertex is at (147/32, 337.64)
On Earth it is about 227.14 feet higher
3) The product of two consecutive odd integers is 1 less than four times their sum.
Find the two integers. 7&9 and 1&-1
We write two consecutive odd integers as:
Let: n = 1st number
n + 2 = 2nd number
Step 1 - Write the equation
Step 2 - Solve the equation
2
n + 2n = 4[2n + 2] - 1
2
n + 2n = 8n + 8 - 1
2
n + 2n = 8n + 7
2
n - 6n - 7= 0
(n - 7)(n + 1) = 0
We see to the right that we get
TWO sets of answers:
7 and 9
- 1 and 1
4) The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches.
Find the dimensions of the rectangle.
Step 1 - Draw the picture of the
rectangle
Step 2 - Write the equation using the
formula LW = A
Step 3 - Solve the equation
Since the length is 6 more than the width we let the
width = x and the length = x + 6
x(x + 6) = 91
2
x + 6x = 91
x + 6x - 91 = 0
(x - 7)(x + 13) = 0
2
The length is 7 and the width is 13
5) The hypotenuse of a right triangle is 6 more than the shorter leg. The longer leg is
three more than the shorter leg. Find the length of the shorter leg. Shorter = 9
Here we use the Pythagorean Theorem which states that in a right triangle:
The sum of the squares of the legs is equal to the square of the hypotenuse.
To work out the problem we can define the sides of the triangle according to the figure below:
Step 1 - Write the equation
Step 2 - Solve the equation
2
2
2
x + (x + 3) = (x + 6)
By using the SQUARE OF A BINOMIAL FORMULA
2
2
2
x + x + 6x + 9 = x + 12x + 36
2
2
2x + 6x + 9 = x + 12x + 36
2
x - 6x - 27 = 0
(x - 9)(x + 3) = 0
6) The length of a rectangle exceeds its width by 3 inches. The area of the
rectangle is 70 square inches, find its dimensions. W = 7 L = 10
Step 1 - Draw the picture of the
rectangle
Since the length is 3 more than the width we let the
width = x and the length = x + 3
Step 2 - Write the equation using the
formula LW = A
x+3
x
x(x + 3) = 70
2
Step 3 - Solve the equation
x + 3x = 70
x + 3x - 70 = 0
(x - 7)(x + 10) = 0
2
The width is 7 and the length is 10
7) The product of two consecutive odd integers is 1 less than twice their sum.
Find the integers. There are two sets of solutions 3,5 and 1, -1
We write two consecutive odd integers as:
Let: n = 1st number
n + 2 = 2nd number
Step 1 - Write the equation
n(n + 2) = 2[n + (n + 2)] – 1
2
w
Step 2 - Solve the equation
2
n + 2n = 2[2n + 2] - 1
2
n + 2n = 4n + 4 - 1
2
n + 2n = 4n + 3
2
n - 2n - 3= 0
(n - 3)(n + 1) = 0
n = 3 and n = 1
We see to the right that we get
TWO sets of answers
3,5 and 1, -1